Question

Show that the points (a, a), (-a, -a) and $\left( { - a\sqrt 3 ,a\sqrt 3 } \right)$ form an equilateral triangle.

Answer 1

Hint: In this question use the distance formula that is $d = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} $ for any two pair of points$\left( {{x_1},{y_1}} \right){\text{ and }}\left( {{x_2},{y_2}} \right)$, and prove that all the sides are of equal length.

Complete step-by-step answer:

The coordinates of the triangle is
A = (a, a) = (x₁, y₁)
B = (-a, -a) = (x₂, y₂)
C = ($ - a\sqrt 3 ,a\sqrt 3 $) = (x₃, y₃)

Proof –
As we know if these coordinates form the equilateral triangle then the length of all sides should be equal.
Now as we know that the distance between two points is calculate as
$d = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} $ So use this property to calculate the length of the sides.
So calculate the sides AB , BC and CA.
$ \Rightarrow AB = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} = \sqrt {{{\left( { - a - a} \right)}^2} + {{\left( { - a - a} \right)}^2}} = \sqrt {4{a^2} + 4{a^2}} = 2a\sqrt 2 $ Units.

Now calculate the distance BC
$ \Rightarrow BC = \sqrt {{{\left( {{x_3} - {x_2}} \right)}^2} + {{\left( {{y_3} - {y_2}} \right)}^2}} = \sqrt {{{\left( { - a\sqrt 3 + a} \right)}^2} + {{\left( {a\sqrt 3 + a} \right)}^2}} $
Now expand the square according to property ${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$ we have,
$ \Rightarrow BC = \sqrt {{{\left( { - a\sqrt 3 + a} \right)}^2} + {{\left( {a\sqrt 3 + a} \right)}^2}} = \sqrt {3{a^2} + {a^2} - 2{a^2}\sqrt 3 + 3{a^2} + {a^2} + 2{a^2}\sqrt 3 } $

Now simplify the above equation we have,
$ \Rightarrow BC = \sqrt {3{a^2} + {a^2} - 2{a^2}\sqrt 3 + 3{a^2} + {a^2} + 2{a^2}\sqrt 3 } = \sqrt {8{a^2}} = 2a\sqrt 2 $ Units.
Now calculate the distance CA
\[ \Rightarrow CA = \sqrt {{{\left( {{x_3} - {x_1}} \right)}^2} + {{\left( {{y_3} - {y_1}} \right)}^2}} = \sqrt {{{\left( { - a\sqrt 3 - a} \right)}^2} + {{\left( {a\sqrt 3 - a} \right)}^2}} \]
Now expand the square according to property ${\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab$ we have,
\[ \Rightarrow CA = \sqrt {{{\left( { - a\sqrt 3 - a} \right)}^2} + {{\left( {a\sqrt 3 - a} \right)}^2}} = \sqrt {3{a^2} + {a^2} + 2{a^2}\sqrt 3 + 3{a^2} + {a^2} - 2{a^2}\sqrt 3 } \]
Now simplify the above equation we have,
$ \Rightarrow CA = \sqrt {3{a^2} + {a^2} + 2{a^2}\sqrt 3 + 3{a^2} + {a^2} - 2{a^2}\sqrt 3 } = \sqrt {8{a^2}} = 2a\sqrt 2 $ Units.
So as we see that all the distances of the sides are equal so the given coordinates form an equilateral triangle.

Hence Proved.

Note: An equilateral triangle is one in which all the three sides are equal, an equilateral triangle is also equiangular that is all the 3 internal angles are congruent to each other and are equal to each other as ${60^0}$. It is advised to remember the distance formula as it helps save a lot of time while solving such problems.